10
$\begingroup$

Can anyone help me find a formal reference for the following identity about the summation of squared tangent function:

$$ \sum_{k=1}^m\tan^2\frac{k\pi}{2m+1} = 2m^2+m,\quad m\in\mathbb{N}^+. $$

I have proved it, however, the proof is too long to be included in a paper. So I just want to refer to some books or published articles.

I also found it to be a special case of the following identity,

$$ \sum_{k=1}^{\lfloor\frac{n-1}{2}\rfloor}\tan^2\frac{k\pi}{n} = \frac16(n-1)(-(-1)^n (n + 1) + 2 n - 1),\quad n\in\mathbb{N}^+ $$

which is provided by Wolfram.

Thank you very much!

$\endgroup$
3
$\begingroup$

Jolley, Summation of Series, formula 445 is $$\sum_{k=0}^{n-1}\tan^2\left(\theta+{k\pi\over n}\right)=n^2\cot^2\left({n\pi\over2}+n\theta\right)+n(n-1)$$ Let $\displaystyle\theta={\pi\over2m+1}$, $n=2m+1$ and we almost have your sum; we have twice your sum, since the angles here go from just over zero to just under $\pi$, while in your sum they go from just over zero to just under $\pi/2$, and $\tan^2\theta=\tan^2(\pi-\theta)$.

Jolley's reference is to page 73 of S L Loney, Plane Trigonometry, Cambridge University Press, 1900. This book is best known from its part in Ramanujan's early education.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

In fact, this type of formula is related to binomial coefficients. I give a proof of the general case I found in my post Tan binomial formulas from a set S and its k-subset

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.